Question

In a leap year, selected at random, find the probability that there are 53 Tuesdays.

Hint:

Remember the number of odd days for different types of years. A normal year (365 days) has 1 odd day (365 = 52 weeks + 1 day), so the probability of 53 of any specific day is \(1/7\). A leap year (366 days) has 2 odd days, so the probability is \(2/7\). This is a common pattern in probability questions.

Answer 1

Step 1: Understanding the Concept:
This question deals with probability. A leap year has 366 days. We need to find the number of weeks and the number of remaining days to determine the probability of having an extra Tuesday.
Step 2: Key Formula or Approach:
Probability is calculated as: \[ P(\text{Event}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} \] Step 3: Detailed Explanation:
A leap year has 366 days.
First, we find the number of complete weeks in 366 days by dividing by 7: \[ 366 \div 7 = 52 \text{ weeks and } 2 \text{ days} \] This means a leap year always contains 52 full weeks. Therefore, there are guaranteed to be 52 Tuesdays.
The occurrence of a 53rd Tuesday depends on the two remaining (odd) days.
These two consecutive days can form the following possible pairs:
(Sunday, Monday)
(Monday, Tuesday)
(Tuesday, Wednesday)
(Wednesday, Thursday)
(Thursday, Friday)
(Friday, Saturday)
(Saturday, Sunday)
There are 7 possible outcomes for the pair of odd days.
For there to be 53 Tuesdays, one of these two odd days must be a Tuesday. The favorable outcomes are the pairs that include a Tuesday:
(Monday, Tuesday)
(Tuesday, Wednesday)
There are 2 favorable outcomes.
Step 4: Final Answer:
The total number of possible outcomes is 7, and the number of favorable outcomes is 2.
Therefore, the probability of a leap year having 53 Tuesdays is: \[ P(53 \text{ Tuesdays}) = \frac{2}{7} \]